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result, we de¬ne f (t) = t 2 +3 and g(t) = 2/t. It follows that r(t) = (g —¦f )(t).

You Try It: Express the function g(x) = 3/(x 2 + 5) as the composition of two
functions. Can you express it as the composition of three functions?

1.8.5 THE INVERSE OF A FUNCTION
Let f be the function which assigns to each working adult American his or her
Social Security Number (a 9-digit string of integers). Let g be the function which
assigns to each working adult American his or her age in years (an integer between
0 and 150). Both functions have the same domain, and both take values in the non-
negative integers. But there is a fundamental difference between f and g. If you
are given a Social Security number, then you can determine the person to whom it
belongs. There will be one and only one person with that number. But if you are
given a number between 0 and 150, then there will probably be millions of people
with that age. You cannot identify a person by his/her age. In summary, if you know
g(x) then you generally cannot determine what x is. But if you know f (x) then you
can determine what (or who) x is. This leads to the main idea of this subsection.
Let f : S ’ T be a function. We say that f has an inverse (is invertible) if
there is a function f ’1 : T ’ S such that (f —¦ f ’1 )(t) = t for all t ∈ T and
(f ’1 —¦ f )(s) = s for all s ∈ S. Notice that the symbol f ’1 denotes a new function
which we call the inverse of f .

Basic Rule for Finding Inverses To ¬nd the inverse of a function f , we
solve the equation
(f —¦ f ’1 )(t) = t
for the function f ’1 (t).
EXAMPLE 1.40
Find the inverse of the function f (s) = 3s.
CHAPTER 1 Basics 43

SOLUTION
We solve the equation
(f —¦ f ’1 )(t) = t.
This is the same as
f (f ’1 (t)) = t.
We can rewrite the last line as
3 · f ’1 (t) = t
or
t
f ’1 (t) = .
3
Thus f ’1 (t) = t/3.
EXAMPLE 1.41
Let f : R ’ R be de¬ned by f (s) = 3s 5 . Find f ’1 .

SOLUTION
We solve
(f —¦ f ’1 )(t) = t
or
f (f ’1 (t)) = t
or
3[f ’1 (t)]5 = t
or
t
[f ’1 (t)]5 =
3
or
1/5
t
’1
(t) =
f .
3

You Try It: Find the inverse of the function g(x) = x ’ 5.
3


It is important to understand that some functions do not have inverses.
EXAMPLE 1.42
Let f : R ’ {t : t ≥ 0} be de¬ned by f (s) = s 2 . If possible, ¬nd f ’1 .
44 CHAPTER 1 Basics

SOLUTION
Using the Basic Rule, we attempt to solve
(f —¦ f ’1 )(t) = t.
Writing this out, we have
[f ’1 (t)]2 = t.
uniquely for f ’1 (t).
But now there is a problem: we cannot solve this equation √

We do not know whether f ’1 (t) = + t or f ’1 (t) = ’ t. Thus f ’1 is not
a well de¬ned function. Therefore f is not invertible and f ’1 does not exist.

Math Note: There is a simple device which often enables us to obtain an inverse”
even in situations like Example 1.42. We change the domain of the function. This
idea is illustrated in the next example.
EXAMPLE 1.43
De¬ne f : {s : s ≥ 0} ’ {t : t ≥ 0} by the formula f (s) = s 2 . Find f ’1 .

SOLUTION
We attempt to solve
(f —¦ f ’1 )(t) = t.
Writing this out, we have
f (f ’1 (t)) = t
or
[f ’1 (t)]2 = t.
This looks like the same situation we had in Example 1.42. But in fact things

have improved. Now we know that f ’1 (t) must be + t, because f ’1 must
have range S = {s : s ≥ 0}. Thus f ’1 : {t : t ≥ 0} ’ {s : s ≥ 0} is given by

f ’1 (t) = + t.

You Try It: The equation y = x 2 +3x does not describe the graph of an invertible
function. Find a way to restrict the domain so that it is invertible.
Now we consider the graph of the inverse function. Suppose that f : S ’ T
is invertible and that (s, t) is a point on the graph of f . Then t = f (s) hence
s = f ’1 (t) so that (t, s) is on the graph of f ’1 . The geometrical connection
between the points (s, t) and (t, s) is exhibited in Fig. 1.47: they are re¬‚ections of
each other in the line y = x. We have discovered the following important principle:
The graph of f ’1 is the re¬‚ection in the line y = x of the graph of f .
Refer to Fig. 1.48.
CHAPTER 1 Basics 45

y




x




Fig. 1.47

y




x




Fig. 1.48

EXAMPLE 1.44
Sketch the graph of the inverse of the function f whose graph is shown in
Fig. 1.49.

SOLUTION
By inspection of the graph we see that f is one-to-one (i.e., takes different
domain values to different range values) and onto (i.e., takes on all values in
the range) from S = [’2, 3] to T = [1, 5]. Therefore f has an inverse. The
graph of f ’1 is exhibited in Fig. 1.50.
46 CHAPTER 1 Basics




Fig. 1.49




Fig. 1.50

You Try It: Sketch f (x) = x 3 + x and its inverse.

Another useful fact is this: Since an invertible function must be one-to-one, two
different x values cannot correspond to (that is, be “sent by the function to”) the
same y value. Looking at Figs. 1.51 and 1.52, we see that this means
In order for f to be invertible, no horizontal line can
intersect the graph of f more than once.




Fig. 1.51
CHAPTER 1 Basics 47

In Fig. 1.51, the fact that the line y = 2 intersects the graph twice means that
the function f takes the value 2 at two different points of its domain (namely at
x = ’2 and x = 6). Thus f is not one-to-one so it cannot be invertible. Figure 1.52
shows what happens if we try to invert f : the resulting curve is not the graph of a
function.




Fig. 1.52

EXAMPLE 1.45
Look at Figs. 1.53 and 1.55. Are the functions whose graphs are shown in
parts (a) and (b) of each ¬gure invertible?




Fig. 1.53

SOLUTION
Graphs (a) and (b) in Fig. 1.53 are the graphs of invertible functions since
no horizontal line intersects each graph more than once. Of course we must
choose the domain and range appropriately. For (a) we take S = [’4, 4] and
T = [’2, 3]; for (b) we take S = (’3, 4) and T = (0, 5). Graphs (a) and (b)
48 CHAPTER 1 Basics

in Fig. 1.54 are the graphs of the inverse functions corresponding to (a) and (b)
of Fig. 1.53 respectively. They are obtained by re¬‚ection in the line y = x.




Fig. 1.54




Y
FL
AM

Fig. 1.55
TE



In Fig. 1.55, graphs (a) and (b) are not the graphs of invertible functions.
For each there is exhibited a horizontal line which intersects the graph twice.
However graphs (a) and (b) in Fig. 1.56 exhibit a way to restrict the domains
of the functions in (a) and (b) of Fig. 1.55 to make them invertible. Graphs (a)
and (b) in Fig. 1.57 show their respective inverses.




Fig. 1.56

You Try It: Give an example of a function from R to R that is not invertible, even
when it is restricted to any interval of length 2.
CHAPTER 1 Basics 49




Fig. 1.57


1.9 A Few Words About Logarithms
and Exponentials
We will give a more thorough treatment of the logarithm and exponential functions
in Chapter 6. For the moment we record a few simple facts so that we may use
these functions in the sections that immediately follow.
The logarithm is a function that is characterized by the property that
log(x · y) = log x + log y.
It follows from this property that
log(x/y) = log x ’ log y
and
log(x n ) = n · log x.
It is useful to think of loga b as the power to which we raise a to get b, for any
a, b > 0. For example, log2 8 = 3 and log3 (1/27) = ’3. This introduces the idea
of the logarithm to a base.

You Try It: Calculate log5 125, log3 (1/81), log2 16.
The most important base for the logarithm is Euler™s number e ≈ 2.71828 . . . .
Then we write ln x = loge x. For the moment we take the logarithm to the base e, or
the natural logarithm, to be given. It is characterized among all logarithm functions
by the fact that its graph has tangent line with slope 1 at x = 1. See Fig. 1.58. Then
we set
ln x
loga x = .
ln a
Note that this formula gives immediately that loge x = ln x, once we accept that
loge e = 1.
50 CHAPTER 1 Basics




Fig. 1.58




Fig. 1.59


Math Note: In mathematics, we commonly write log x to mean the natural log-
arithm. Thus you will sometimes encounter ln x and sometimes encounter log x
(without any subscript); they are both understood to mean loge x, the natural
logarithm.
The exponential function exp x is de¬ned to be the inverse function to ln x.
Figure 1.59 shows the graph of y = exp x. In fact we will see later that exp x = ex .
More generally, the function a x is the inverse function to loga x. The exponential
has these properties:
a b+c = a b · a c ;
(a)
CHAPTER 1 Basics 51

(a b )c = a b·c ;
(b)
ab
= c.
b’c
(c) a
a
These are really just restatements of properties of the logarithm function that we
have already considered.

You Try It: Simplify the expressions 32 · 54 /(15)3 and 24 · 63 · 12’4 .




Exercises
1. Each of the following is a rational number. Write it as the quotient of two
integers.
2/3 ’ 7/8
(a)
(b) 43.219445
’37 ’4
·
(c)
533 ’6
2
(d)
3.45969696 . . .
’73.235677677677 . . .
(e)
3
5
(f) ’17
4 +9
3

’4
9 +5
2
(g) ’11
3 +7
6


(h) 3.2147569569569 . . .
√ √
2. Plot the numbers 3.4, ’π/2, 2π , ’ 2 + 1, 3 · 4, 9/2, ’29/10 on a real
number line. Label each plotted point.
3. Sketch each of the following sets on a separate real number line.
S = {x ∈ R: |x ’ 2| < 4}
(a)
T = {t ∈ R: t 2 + 1 = 5}
(b)
U = {s ∈ R: 2s ’ 5 ¤ 3}
(c)
V = {y ∈ R: |6y + 1| > 2}
(d)
S = {x ∈ R: x 2 + 3 < 6}
(e)
T = {s ∈ R: |s| = |s + 1|}
(f )
52 CHAPTER 1 Basics

√√ √
4. Plot each of the points (2, ’4), (’6, 3), (π, π 2 ), (’ 5, 8), ( 2π, ’3),
(1/3, ’19/4) on a pair of cartesian coordinate axes. Label each point.
5. Plot each of these planar loci on a separate set of axes.
{(x, y): y = 2x 2 ’ 3}
(a)
{(x, y): x 2 + y 2 = 9}
(b)
y = x3 + x
(c)
x = y3 + y
(d)
x = y2 ’ y3
(e)
x2 + y4 = 3
(f)
6. Plot each of these regions in the plane.
{(x, y): x 2 + y 2 < 4}
(a)
{(x, y): y > x 2 }
(b)
{(x, y): y < x 3 }
(c)
{(x, y): x ≥ 2y + 3}
(d)
{(x, y): y ¤ x + 1}
(e)
{(x, y): 2x + y ≥ 1}
(f)
7. Calculate the slope of each of the following lines:
(a) The line through the points (’5, 6) and (2, 4)
(b) The line perpendicular to the line through (1, 2) and (3, 4)
The line 2y + 3x = 6
(c)
x ’ 4y
=6
(d) The line
x+y
(e) The line through the points (1, 1) and (’8, 9)
The line x ’ y = 4
(f)
8. Write the equation of each of the following lines.
The line parallel to 3x + 8y = ’9 and passing through the point
(a)
(4, ’9).
The line perpendicular to x + y = 2 and passing through the point
(b)
(’4, ’8).
The line passing through the point (4, 6) and having slope ’8.
(c)
(d) The line passing through (’6, 4) and (2, 3).
(e) The line passing through the origin and having slope 6.
The line perpendicular to x = 3y ’ 7 and passing through (’4, 7).
(f)

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